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I have the following complex equation:

\begin{equation} z^6 + 1 = 0 \end{equation}

I would like to be able to gain some intuition and understanding. I know from the fundamental theorem of algebra that any polynomial with degree n has n roots. So it is plausible to say that the above equation has six roots. However, what about repeated roots?

Since I am not experienced with complex equations, to me all I see is that $\pm i$ are two solutions, but are there others?

The CAS Sage gave me the following output (though I never know when to trust it):

[z == 1/2*I*(-1)^(1/6)*sqrt(3) + 1/2*(-1)^(1/6), z == 1/2*I*(-1)^(1/6)*sqrt(3) - 1/2*(-1)^(1/6), z == -(-1)^(1/6), z == -1/2*I*(-1)^(1/6)*sqrt(3) - 1/2*(-1)^(1/6), z == -1/2*I*(-1)^(1/6)*sqrt(3) + 1/2*(-1)^(1/6), z == (-1)^(1/6)]

Thanks for all the help in advance

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6 Answers 6

up vote 2 down vote accepted

They're all roots of $-1$, which has absolute value $1$, so they all have absolute value $1$ as well. That means they're found on the unit circle in $\mathbb{C}$.

As for how they're distributed around the circle, if you measure angles from the positive real axis, you want angles which, when multiplied by $6$ give you $\pi$, so you're looking at odd multiples of $\frac{\pi}{6}$. That includes $\frac{3\pi}{6}$ and $\frac{9\pi}{6}$, which correspond to the answers $\pm i$ that you noted. The other $4$ roots are $\pm \frac{\sqrt{3}}{2}\pm \frac{1}{2}i$, where the two plus/minus symbols can vary independently, giving $4$ values.

Does this help?

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Yes! However, can you explain the "multiply by 6." I think this is the definition of the argument... –  CodeKingPlusPlus Sep 6 '13 at 4:18
    
Yeah, the "argument" is the angle from the positive real axis. When you multiply two complex numbers, you add the arguments, so raising one to the sixth power multiplies its argument by $6$. The modulus (or absolute value) is multiplied when the numbers are multiplied, so we have $1^6=1$, and that's fine. –  G Tony Jacobs Sep 6 '13 at 4:47

Are you familiar with the polar form for complex numbers. You have $z^6=-1$ If you write $z=re^{i\theta},z^6=r^6e^{6i\theta}$ What does this tell you about $r, \theta?$

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Since $e^{i2\pi k}=1,\, k\in\mathbb{Z}$, we have

$$ z^6 + 1 = 0 \implies z^6=-1=e^{i\pi} e^{i2\pi k}=e^{ i(2k+1)\pi } \implies z= e^\frac{{ i(2k+1)\pi }}{6}, k=0,1,2,3,4,5.$$

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$$z^6=-1=\cos\pi=\text{cis}{\pi}$$

Then,

$$z_k=\text{cis}{\frac{\pi+2k\pi}{6}}$$

the six roots obtained with setting $k=0,1,2,3,4,5.$

Further reading: Paul's Online Math Notes - Roots of complex numbers

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z^6 = -1
In polar form:
(re^(i theta))^6 = -1
r^6 e^(6 i theta) = e^(i (pi + 2*pi*k))
r = 1.
e^(6 i theta) = e^(i (pi + 2*pi*k))
6 i theta = i (pi + 2*pi*k)
6 theta = pi + 2*pi*k
theta = (2k+1) pi / 6.
Then
z = cos ((2k+1) pi / 6) + i sin((2k+1) pi / 6).
Different ones are produced when k = 0,1,2,3,4,5.
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It's true. A good group of mathematicians don't give any credit. We solve equations like $z^6$+1=0 by use of De'Moivre's formula. The history behind such problems like this provide a better understanding of such an unnatural solution. How can a number raised to an even power plus another even number ever be zero? We also need to pay attention to the fact that a n-degree polynomial will give us n solutions and will form an n-gon in the complex plane. Also, remember cos, sin are 2$\pi$ periodic, hence adding 2$\pi$k.

All other algebra done in previous posts are correct, i just wanted to give some insight.

http://en.wikipedia.org/wiki/De_Moivre%27s_formula

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